{"id":40727,"date":"2020-02-24T18:41:35","date_gmt":"2020-02-24T13:11:35","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40727"},"modified":"2020-02-24T18:42:15","modified_gmt":"2020-02-24T13:12:15","slug":"height-and-distance-questions-for-ssc-chsl-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/height-and-distance-questions-for-ssc-chsl-set-2-pdf\/","title":{"rendered":"Height and Distance Questions for SSC CHSL set-2 PDF"},"content":{"rendered":"

Height and Distance Questions for SSC CHSL set-2 PDF<\/strong><\/span><\/h1>\n
Download SSC CHSL Height and Distance questions with answers Set-2 PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Height and Distance Questions for SSC Exam<\/h5>\n

Download Height and Distance Questions for SSC CHSL set-2 PDF<\/a><\/p>\n

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Take a free mock test for SSC CHSL<\/a><\/p>\n

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More SSC CHSL Important Questions and Answers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>From an aeroplane just over a straight road, the angles of depression of two consecutive kilometre stones situated at opposite sides of the aeroplane were found to be 60\u00b0 and 30\u00b0 respectively. The height (in km) of the aeroplane from the road at that instant was (Given \u221a3 = 1.732)<\/p>\n

a)\u00a00.433<\/p>\n

b)\u00a08.66<\/p>\n

c)\u00a04.33<\/p>\n

d)\u00a00.866<\/p>\n

Question 2:\u00a0<\/b>A person of height 6ft. wants to pluck a fruit which is on a 26\/3 ft. high tree. If the person is standing 8\/\u221a3 ft. away from the base of the tree, then at what angle should he throw a stone so that it hits the fruit ?<\/p>\n

a)\u00a075\u00b0<\/p>\n

b)\u00a030\u00b0<\/p>\n

c)\u00a045\u00b0<\/p>\n

d)\u00a060\u00b0<\/p>\n

Question 3:\u00a0<\/b>The angle of elevation of a tower from a distance of 100 metre from its foot is 30\u00b0. Then the height of the tower is<\/p>\n

a)\u00a0$50\\sqrt{3}$ metre<\/p>\n

b)\u00a0$100\\sqrt{3}$ metre<\/p>\n

c)\u00a0$\\frac{50}{\\sqrt{3}}$ metre<\/p>\n

d)\u00a0$\\frac{100}{\\sqrt{3}}$ metre<\/p>\n

Question 4:\u00a0<\/b>A kite is flying at the height of 75m from the ground. The string makes an angle \u03b8 (where cot\u03b8 = 8\/15) with the level ground. Assuming that there is no slack in the string the length of the string is equal to :<\/p>\n

a)\u00a085 metre<\/p>\n

b)\u00a065 metre<\/p>\n

c)\u00a075 metre<\/p>\n

d)\u00a040 metre<\/p>\n

Question 5:\u00a0<\/b>The angle of elevation of the top of a vertical tower situated perpendicularly on a plane is observed as 60\u00b0 from a point P on the same plane. From another point Q, 10m vertically above the point P, the angle of depression of the foot of the tower is 30\u00b0. The height of the tower is<\/p>\n

a)\u00a015 m<\/p>\n

b)\u00a030 m<\/p>\n

c)\u00a020 m<\/p>\n

d)\u00a025 m<\/p>\n

SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n\n

SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n

Question 6:\u00a0<\/b>From a point 20 m away from the foot of a tower, the angle of elevation of the top of the tower is 30\u00b0. The height of the tower is<\/p>\n

a)\u00a010 \u221a3 m<\/p>\n

b)\u00a020 \u221a3 m<\/p>\n

c)\u00a0$\\frac{10}{\\sqrt{3}}$ m<\/p>\n

d)\u00a0$\\frac{20}{\\sqrt{3}}$ m<\/p>\n

Question 7:\u00a0<\/b>The length of the shadow of a tower is 9 metres when the sun\u2019s altitude is 30\u00b0. What is the height of the tower ?<\/p>\n

a)\u00a0$3\\sqrt{3}$ m<\/p>\n

b)\u00a0$4\\frac{1}{2}$ m<\/p>\n

c)\u00a0$9\\sqrt{3}$ m<\/p>\n

d)\u00a0$\\frac{9\\sqrt{3}}{2}$ m<\/p>\n

Question 8:\u00a0<\/b>A right pyramid 6 m height has a square base of which the diagonal is $\\sqrt{1152} m$. Volume of the pyramid is<\/p>\n

a)\u00a0144 m<\/p>\n

b)\u00a0288 m<\/p>\n

c)\u00a0576 m<\/p>\n

d)\u00a01152 m<\/p>\n

Question 9:\u00a0<\/b>From the top of a cliff 90 metre high, the angles of depression of the top and bottom of a tower are observed to be 30\u00b0 and 60\u00b0 respectively. The height of the tower is :<\/p>\n

a)\u00a045 m<\/p>\n

b)\u00a060 m<\/p>\n

c)\u00a075 m<\/p>\n

d)\u00a030 m<\/p>\n

Question 10:\u00a0<\/b>A boy standing in the middle of a field, observes a flying bird in the north at an angle of elevation of 30\u00b0 and after 2 minutes, he observes the same bird in the south at an angle of elevation of 60\u00b0. If the bird flies all along in a straight line at a height of 50 m, then its speed in km\/h is :<\/p>\n

a)\u00a04.5<\/p>\n

b)\u00a03<\/p>\n

c)\u00a09<\/p>\n

d)\u00a06<\/p>\n

Question 11:\u00a0<\/b>A tree is broken by the wind. If the top of the tree struck the ground at an angle of 30\u00b0 and at a distance of 30 m from the root, then the height of the tree is<\/p>\n

a)\u00a025\u221a3 m<\/p>\n

b)\u00a030\u221a3 m<\/p>\n

c)\u00a015\u221a3 m<\/p>\n

d)\u00a020\u221a3 m<\/p>\n

Question 12:\u00a0<\/b>At an instant, the length of the shadow of a pole is , \u221a3 times the height of the pole. The angle of elevation of the Sun at that moment is<\/p>\n

a)\u00a075\u00b0<\/p>\n

b)\u00a030\u00b0<\/p>\n

c)\u00a045\u00b0<\/p>\n

d)\u00a060\u00b0<\/p>\n

Question 13:\u00a0<\/b>The height of a tower is h and the angle of elevation of the top of the tower is a. On moving a distance h\/2 towards, the tower, the angle of elevation becomes 0. The value of cot\u03b1 – cot \u03b2 is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a01\/2<\/p>\n

d)\u00a02\/3<\/p>\n

Question 14:\u00a0<\/b>If the angles of elevation of a balloon from two consecutive kilometre-stones along a road are 30\u00b0 and 60\u00b0 respectively, then the height of the balloon above the
\nground will be<\/p>\n

a)\u00a0$\\frac{\\sqrt{3}}{2}$ km<\/p>\n

b)\u00a0$\\frac{1}{2}$ km<\/p>\n

c)\u00a0$\\frac{2}{\\sqrt{3}}$ km<\/p>\n

d)\u00a0$3\\sqrt{3}$<\/p>\n

Question 15:\u00a0<\/b>A tower is broken at a point P above the ground. The top of the tower makes an angle 60\u00b0 with the ground at Q. From another point R on the opposite side of Q angle of elevation of point P is 30\u00b0. If QR = 180 m, then what is the total height (in metres) of the tower?<\/p>\n

a)\u00a0$90$<\/p>\n

b)\u00a0$45\\sqrt{3}$<\/p>\n

c)\u00a0$45(\\sqrt{3}+1)$<\/p>\n

d)\u00a0$45(\\sqrt{3}+2)$<\/p>\n

SSC CHSL FREE MOCK TEST<\/a><\/p>\n

FREE SSC STUDY MATERIAL<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

<\/p>\n

OC = height of plane = $h$<\/p>\n

$\\angle$OAC = $\\angle$DOA = 60\u00b0<\/p>\n

$\\angle$OBC = $\\angle$BOE = 30\u00b0<\/p>\n

AB = 2 and let AC = $x$<\/p>\n

=> BC = $(2-x)$<\/p>\n

From, $\\triangle$OAC<\/p>\n

$tan60^{\\circ} = \\frac{OC}{AC}$<\/p>\n

=> $\\sqrt{3} = \\frac{h}{x}$<\/p>\n

=> $x = \\frac{h}{\\sqrt{3}}$ ————Eqn(1)<\/p>\n

From, $\\triangle$OBC<\/p>\n

$tan30^{\\circ} = \\frac{OC}{BC}$<\/p>\n

=> $\\frac{1}{\\sqrt{3}} = \\frac{h}{2-x}$<\/p>\n

=> $\\sqrt{3}h = 2 – \\frac{h}{\\sqrt{3}}$ [From eqn(1)]<\/p>\n

=> $\\frac{3h+h}{\\sqrt{3}} = 2$<\/p>\n

=> $h = \\frac{2\\sqrt{3}}{4} = \\frac{\\sqrt{3}}{2}$<\/p>\n

= $\\frac{1.732}{2}$ = 0.866<\/strong><\/p>\n

2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

<\/p>\n

Height of person = CD = 6 ft<\/p>\n

Height of tree = AB = $\\frac{26}{3}$ ft<\/p>\n

Distance between them = BD = $\\frac{8}{\\sqrt{3}}$ ft<\/p>\n

To find : $\\angle$ACE = $\\theta$ = ?<\/p>\n

Solution : AE = AB – BE = $\\frac{26}{3}$ – 6<\/p>\n

=> AE = $\\frac{8}{3} ft$<\/p>\n

and BD = CE = $\\frac{8}{\\sqrt{3}}$ ft<\/p>\n

Now, in $\\triangle$AEC<\/p>\n

=> $tan\\theta$ = $\\frac{AE}{CE}$<\/p>\n

=> $tan\\theta$ = $\\frac{\\frac{8}{3}}{\\frac{8}{\\sqrt{3}}}$<\/span><\/p>\n

=> $tan\\theta$ = $\\frac{1}{\\sqrt{3}}$<\/span><\/span><\/p>\n

=> $\\theta$ = 30\u00b0<\/span><\/span><\/span><\/p>\n

3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

<\/p>\n

Height of tower = AB<\/p>\n

In $\\triangle$ABC<\/p>\n

=> $tan\\theta = \\frac{AB}{BC}$<\/p>\n

=> $tan30^{\\circ} = \\frac{AB}{100}$<\/p>\n

=> $\\frac{1}{\\sqrt{3}} = \\frac{AB}{100}$<\/p>\n

=> $AB = \\frac{100}{\\sqrt{3}}$<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

<\/p>\n

Height of kite from ground = AB = 75 m<\/p>\n

$\\angle$ACB = $\\theta$<\/p>\n

We know that $cot\\theta = \\frac{8}{15}$<\/p>\n

=> $\\frac{BC}{AB} = \\frac{8}{15}$<\/p>\n

=> $BC = \\frac{8*75}{15} = 40 m$<\/p>\n

Now, length of string AC = $\\sqrt{(AB)^2 + (BC)^2}$<\/p>\n

=> AC = $\\sqrt{75^2 + 40^2}$<\/p>\n

= $\\sqrt{5625+1600} = \\sqrt{7225}$<\/p>\n

=> AC = 85 m<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

In the given problem, distance between tower and point of observation will be the adjacent side and tower will be the opposite side.
\nLet ‘d’ be the distance and ‘h’ be the height of the tower.
\ntan 60 = h\/d => d = h\/tan 30
\ntan 30 = (h-10)\/d => d = (h-10)\/ tan 60
\nh\/tan 30 = (h-10)\/tan 60
\nh\/h-10 = 3
\n2h = 30
\nh = 15m
\nOption A is the correct answer.<\/p>\n

6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Perpendicular AB= height of tower<\/p>\n

Base BC = distance b\/w base of tower and the point<\/p>\n

Hypotenuse AC= segment b\/w top of the tower and the point<\/p>\n

Angle of elevation ACB =30\u00b0<\/p>\n

tan 30\u00b0 = Perpendicular \/ Base = 1\/\u221a3<\/p>\n

Perpendicular = Base*1\/\u221a3<\/p>\n

Base is 20 m.<\/p>\n

So height of tower is 20\/\u221a3m.<\/p>\n

7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Length of shadow will be the adjacent side.<\/p>\n

Height of tower will be the opposite side.
\nHence, tan 30 = height\/length
\n$1\/\\sqrt{3}$ = $h\/9$
\n=>$h = 3\\sqrt{3}$
\nOption A is the right answer.<\/span><\/p>\n

8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

It is given that base of pyramid is square and diagonal is given as $\\sqrt{1152} m$<\/p>\n

we know diagonal of sqaure is $\\sqrt{2}$ a , where a is the side of sqaure<\/p>\n

so, $\\sqrt{1152} m$ = $\\sqrt{2} a$<\/span><\/p>\n

a = 24 m<\/span><\/p>\n

Area of base = $\\text{side}^2$ = $24^2$ = 576 <\/span><\/p>\n

Height of pyramid = 6 m <\/span><\/p>\n

Volume of right pyramid = $\\frac{1}{3} {\\times \\text{Base Area} \\times \\text{Height}}$ = $\\frac{1}{3} \\times 576 \\times 6$ = $1152m^{3}$<\/span><\/span><\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

<\/p>\n

Given: CD = 90 meter;
\n\u2220KBD = 30\u00b0 & \u2220CAD = 60\u00b0
\nBD \u00d7 cos30\u00b0 = BK
\nAD \u00d7 cos60\u00b0 = AC
\nBD \u00d7 sin30\u00b0 = DK
\nAD \u00d7 sin60\u00b0 = DC
\nBK = AC
\nBD \u00d7 cos30\u00b0 = AD \u00d7 cos60\u00b0
\n\u221a 3 \u00d7 BD = AD ____________ (1)
\nNow, DC = DK + KC
\nDC – DK = AB [As, KC = AB]
\n(AD \u00d7 sin60\u00b0) – (BD \u00d7 sin30\u00b0) = AB
\nUsing equation (1)
\n($\\sqrt{3}$ <\/span>\u00d7<\/span>B<\/span>D<\/span>\u00d7$\\frac{\\sqrt{3}}{2}$<\/span><\/span><\/span>)<\/span><\/span>\u2212<\/span>B<\/span>D<\/span><\/span><\/span>2<\/span><\/span>=<\/span>A<\/span>B<\/span><\/span><\/span><\/p>\n

(3\u00d7BD\u00d732)\u2212BD2=AB<\/p>\n

AB = BD _____________ (2)
\nNow, AD \u00d7 sin 60\u00b0 = DC = 90 meter
\nA<\/span>D<\/span>=$\\frac{180}{\\sqrt{3}}$<\/span><\/span><\/span><\/p>\n

From (1)
\nB<\/span>D<\/span>=<\/span>A<\/span>D\/$\\sqrt{3}$<\/span><\/span><\/span><\/span>
\n<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

=180\/$\\sqrt{3} \\times \\sqrt{3} $ = <\/span>60 meter<\/p>\n

From (2) AB = 60 meter
\nOption B is the correct answer.<\/p>\n

10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n


\nFrom the diagram,
\nHeight = AD = 50\u221a3 m
\n\u2220BAN = 30\u00b0
\n\u2220CAM = 60\u00b0
\n\u2234\u2220BAD = 90\u00b0 – 30\u00b0 = 60\u00b0
\n\u2234\u2220CAD = 90\u00b0 – 60\u00b0 = 30\u00b0
\nFrom \u0394ABD,
\ntan\u2220BAD = Perpendicular\/ Base
\ntan60\u00b0 = BD\/AD
\n\u221a3 = BD\/(50\u221a3)
\nBD = 50 \u00d7 3 = 150 m
\nFrom \u0394ACD,
\ntan\u2220CAD = Perpendicular\/ Base
\ntan30\u00b0 = CD\/AD
\n1\/\u221a3 = CD\/(50\u221a3)
\nCD = 50 m
\n\u2234 Distance travelled by the bird
\n= BC = BD + CD = 150 m + 50 m = 200 m = 0.200 km
\nTime taken to cover this distance = 2 minutes = 2\/60 hr = 1\/30 hr
\n\u2234 Speed
\n= Distance travelled\/ Time required
\n=<\/span>0.200 $\\times$<\/span><\/span><\/span>30<\/span><\/span><\/span><\/span><\/span><\/span><\/span>k<\/span>m<\/span>\/<\/span><\/span><\/span>h<\/span>r<\/span><\/span><\/span><\/p>\n

= 0.200 \u00d7 30 km\/hr
\n= 6 km\/hr
\nOption D is the correct answer<\/p>\n

11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n


\n$tan 30 = \\frac{AB}{BC}$ =$ \\frac{1}{\\sqrt{3}}$
\n$cos 30 = \\frac{BC}{AC} = \\frac{sqrt{3}}{2}$
\nHeight of the tree = $AB + AC$
\n$AB= BC \\times \\frac{1}{\\sqrt{3}}$
\n$AB= 30 \\times \\frac{1}{\\sqrt{3}} = \\frac{30}{\\sqrt{3}}$
\n$AC = \\frac {2 \\times BC}{\\sqrt{3}} =\\frac {2 \\times 30}{\\sqrt{3}}$
\n$AB+AC = \\frac{30}{\\sqrt{3}} + \\frac {60}{\\sqrt{3}} = 30\\sqrt{3}$
\nHence Option B is the correct answer.<\/span><\/span><\/p>\n

12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

<\/p>\n

Let height of pole = AB = $h$<\/p>\n

Length of shadow = BC = $x$<\/p>\n

Angle of elevation of sun = $\\angle$ACB = $\\theta$ = ?<\/p>\n

Acc to ques : $x = \\sqrt{3} h$<\/p>\n

In $\\triangle$ABC<\/p>\n

=> $tan \\theta = \\frac{AB}{BC}$<\/p>\n

=> $tan \\theta = \\frac{h}{x}$<\/p>\n

=> $tan \\theta = \\frac{h}{\\sqrt{3} h} = \\frac{1}{\\sqrt{3}}$<\/p>\n

=> $\\theta$ = 30\u00b0<\/p>\n

13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/p>\n

Here, $\\angle$ACB = $\\alpha$ and $\\angle$ADB = $\\beta$<\/p>\n

AB = tower = $h$ metre<\/p>\n

and CD = $\\frac{h}{2}$ metre<\/p>\n

From $\\triangle$ABC<\/p>\n

=> $tan \\alpha = \\frac{AB}{BC} = \\frac{h}{BC}$<\/p>\n

=> $BC = h cot \\alpha$ ———-Eqn(1)<\/p>\n

From $\\triangle$ABD<\/p>\n

=> $tan \\beta = \\frac{AB}{BD} = \\frac{h}{BC – CD}$<\/p>\n

=> $tan \\beta = \\frac{h}{h cot \\alpha – \\frac{h}{2}}$<\/p>\n

=> $h cot \\alpha – \\frac{h}{2} = h cot \\beta$<\/p>\n

=> $h (cot \\alpha – cot \\beta) = \\frac{h}{2}$<\/p>\n

=> $cot \\alpha – cot \\beta = \\frac{1}{2}$<\/p>\n

14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

<\/p>\n

It is given that BC = 2 km<\/p>\n

Let BD = x => CD = (2 – x) km<\/p>\n

From $\\triangle$ABD<\/p>\n

=> $tan 30 = \\frac{AD}{BD}$<\/p>\n

=> $\\frac{1}{\\sqrt{3}} = \\frac{AD}{x}$<\/p>\n

=> $AD = \\frac{x}{\\sqrt{3}}$<\/p>\n

From $\\triangle$ADC<\/p>\n

=> $tan 60 = \\frac{AD}{CD}$<\/p>\n

=> $\\sqrt{3} = \\frac{\\frac{x}{\\sqrt{3}}}{2 – x}$<\/p>\n

=> $3 (2 – x) = x$<\/p>\n

=> $x = \\frac{3}{2}$<\/p>\n

Now, height of balloon above ground = $AD = \\frac{\\frac{3}{2}}{\\sqrt{3}}$<\/p>\n

= $\\frac{\\sqrt{3}}{2}$ km<\/p>\n

15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

<\/figure>\n

In $\\triangle$ PRS,<\/p>\n

=> $tan(30^\\circ)=\\frac{PS}{RS}$<\/p>\n

=> $\\frac{1}{\\sqrt3}=\\frac{x}{d}$<\/p>\n

=> $d=\\sqrt3x$ ————(i)<\/p>\n

Similarly, in $\\triangle$ PQS,<\/p>\n

=> $tan(60^\\circ)=\\frac{PS}{SQ}$<\/p>\n

=> $\\sqrt3=\\frac{x}{180-d}$<\/p>\n

=> $180\\sqrt3-3x=x$ \u00a0 \u00a0 [Using equation (i)]<\/p>\n

=> $x+3x=4x=180\\sqrt3$<\/p>\n

=> $x=\\frac{180\\sqrt3}{4}=45\\sqrt3$ ————(ii)<\/p>\n

Again, in $\\triangle$ PQS,<\/p>\n

=> $sin(60^\\circ)=\\frac{PS}{PQ}$<\/p>\n

=> $\\frac{\\sqrt3}{2}=\\frac{x}{y}$<\/p>\n

=> $\\sqrt3y=2(45\\sqrt3)$ \u00a0 \u00a0 \u00a0\u00a0[Using equation (ii)]<\/p>\n

=> $y=\\frac{90\\sqrt3}{\\sqrt3}=90$ ———–(iii)<\/p>\n

Adding equations (ii) and (iii), we get\u00a0:<\/p>\n

=> $x+y=45\\sqrt3+90$<\/p>\n

=> Height of tower = $45(\\sqrt3+2)$ m<\/p>\n

=> Ans – (D)<\/p>\n

DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n

SSC CHSL Important Q&A PDF<\/a><\/p>\n

We hope this Height and Distance\u00a0 questions of set-2 pdf for SSC CHSL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Height and Distance Questions for SSC CHSL set-2 PDF Download SSC CHSL Height and Distance questions with answers Set-2 PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Height and Distance Questions for SSC Exam Take a free mock test for SSC CHSL Download SSC CHSL Previous Papers More SSC […]<\/p>\n","protected":false},"author":49,"featured_media":40732,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1493,1459,1611,1741,1441],"tags":[3555,358],"class_list":{"0":"post-40727","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-gd","14":"category-ssc-je","15":"category-ssc-mts","16":"category-stenographer","17":"tag-height-and-distance-questions","18":"tag-ssc-chsl"},"better_featured_image":{"id":40732,"alt_text":"Height and Distance Questions for SSC CHSL set-2 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